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Let $X$ be a scheme. What are the compact objects in the category of quasi-coherent $\mathcal{O}_X$-modules? All references seem to discuss the derived category but I need the abelian category.

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$\DeclareMathOperator\colim{colim}\DeclareMathOperator\Qch{Qch}\DeclareMathOperator\Coh{Coh}\DeclareMathOperator\Hom{Hom}\newcommand\Ab{\mathrm{Ab}}\newcommand\Id{\mathrm{Id}}$In a category admitting small filtrant inductive limits, we call an object $X$ compact if $\Hom(X,\cdot)$ commutes with small filtrant inductive limits. Let $X$ be a Noetherian scheme. Then $F$ is a compact object of $\Qch(O_X)$ iff $F\in \Coh(O_X)$.

Proof (similar to that of https://stacks.math.columbia.edu/tag/0G8P) learned from a friend: Let $\colim_iG_i$ be a filtered colimit in $\Qch(O_X)$.

Step 1. We prove that $O_X$ is compact: Because the underlying topology of $X$ is locally Noetherian, by Stacks 01FF, the canonical map $\colim_i\Gamma(X,G_i)\to \Gamma(X,\colim_iG_i)$ is an isomorphism.

Step 2. If $O_X^n\to O_X^m\to F\to0$ is an exact sequence, then there is a commutative diagram with exact rows $$\require{AMScd}\begin{CD} 0 @>>> \colim_i \Hom(F, G_i) @>>> \colim_i \Hom(O_X^m, G_i) @>>> \colim_i \Hom(O_X^n, G_i) \\ @. @VVV @VVV @VVV \\ 0 @>>> \Hom(F, \colim_i G_i) @>>> \Hom(O_X^m, \colim_i G_i) @>>> \Hom(O_X^n, \colim_i G_i) \end{CD}$$ (TikzCD link). The two vertical homomorphisms on the right are isomorphisms, so is the vertical one on the left.

Step 3. Assume that $F\in \Coh(O_X)$, then there is an open cover $\{U_{\alpha}\}_{\alpha}$ of $X$ s.t. for every $\alpha$, there is an exact sequence $O_X^{n_{\alpha}}\rvert_{U_{\alpha}}\to O_X^{m_{\alpha}}\rvert_{U_{\alpha}}\to F\rvert_{U_{\alpha}}$. Since the category $\Ab$ satisfies AB5, $\colim_i\Hom(F,G_i)$ is the equalizer of the two natural maps $\prod_{\alpha}\colim_i\Hom(F\rvert_{U_{\alpha}},G_i)\rightrightarrows \prod_{\alpha,\beta}\colim_i\Hom(F\rvert_{U_{\alpha}\cap U_{\beta}},G_i)$. The following commutative diagram has exact rows $$\begin{CD} 0 @>>> \colim_i \Hom(F, G_i) @>>> \prod_\alpha \colim_i \Hom(F\rvert_{U_\alpha}, G_i) @>>> \prod_{\alpha, \beta} \colim_i \Hom(F\rvert_{U_\alpha \cap U_\beta}, G_i) \\ @. @VVV @VVV @VVV \\ 0 @>>> \Hom(F, \colim_i G_i) @>>> \prod_\alpha \Hom(F\rvert_{U_\alpha}, G_i) @>>> \prod_{\alpha, \beta} \Hom(F\rvert_{U_\alpha \cap U_\beta}, \colim_i G_i) \end{CD}$$ (TikzCD link). By last paragraph, the natural map $\colim_i\Hom(F|_{U_{\alpha}},G_i\rvert_{U_{\alpha}})\to \Hom(F\rvert_{U_{\alpha}},(\colim_iG_i)\rvert_{U_{\alpha}})$ is an isomorphism. Hence, the two vertical homomorphisms on the right are isomorphisms, so is the vertical one on the left. This proves that $F$ is compact.

Step 4. Conversely, assume that $F$ is a compact object. By a result of Deligne (Proposition 2, p.407 of Residues and duality by Hartshorne), $F=\colim_iF_i$ is a filtrant colimit of a family of coherent modules $\{F_i\}$. Then $\Hom(F,F)=\Hom(F,\colim_i F_i)=\colim_i\Hom(F,F_i)$, so there is $i_0\in I$ and $\iota:F\to F_{i_0}$ st $r\iota=\Id_F$, where $r:F_{i_0}\to F$ is the canonical morphism. Then $r$ is surjective, so $F$ is finite type. For every open subset $U$ of $X$, every $n\ge 1$, the kernel of $f:O_X^n\rvert_U\to F\rvert_U$ is the kernel of $\iota f:O_X^n\rvert_U\to F_{i_0}\rvert_U$. Then the kernel has finite type. Thus, $F$ is coherent.

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    $\begingroup$ @LSpice, Your efforts and time are much appreciate. I am not good at adding links and diagrams. Thank you for correcting my mistakes! $\endgroup$
    – Doug Liu
    Jul 20, 2023 at 19:22

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